Answer to **Puzzle 196**: Let's call our monoidal poset \$$\mathcal{P}\$$. For every \$$a\in\mathcal{P}\$$ such that \$$a\leq x\multimap y\$$, \$$a\leq x\multimap y\Leftrightarrow x\otimes a\leq y\$$. If \$$a,b\in\mathcal{P}\$$ and \$$x\otimes a\leq y\$$ and \$$x\otimes b\leq y\$$, then \$$a\leq x\multimap y\$$ and \$$b\leq x\multimap y\$$, so \$$x\multimap y\$$ is an upper bound in \$$\mathcal{P}\$$ for \$$a\$$ and \$$b\$$. By the same argument, \$$x\multimap y\$$ must be an upper bound for \$$\\{a\in\mathcal{P}:x\otimes a\leq y\\}\$$, i.e., \$\bigvee \lbrace a : \; x \otimes a \le y \rbrace\leq x\multimap y\$ . Since \$$x\multimap y\in\mathcal{P}\$$, and \$$x\multimap y\leq x\multimap y\$$ so that \$$x\otimes(x\multimap y)\leq y\$$, \$$x\multimap y\$$ is an element of the set of which it is an upper bound. Therefore it is the least upper bound, i.e., \$x\multimap y=\bigvee \lbrace a : \; x \otimes a \le y \rbrace\$ .

Answer to **Puzzle 194**: \$$S\multimap T=\bigvee \lbrace A : \; S \otimes A \le T \rbrace=\bigvee \lbrace A : \; S \cup A \subseteq T\rbrace\$$, so this is the set with the most elements such that its union with \$$S\$$ is \$$T\$$. This set is \$$S^c\cap T\$$ (where \$$S^c\$$ denotes the set complement operation), since the addition of any more elements of \$$X\$$ would make its union with \$$S\$$ a proper superset of \$$T\$$.